Question: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}-8x-8y &= 1 \\ -x+6y &= -6\end{align*}$
Begin by moving the $x$ -term in the second equation to the right side of the equation. $6y = x-6$ Divide both sides by $6$ to isolate $y$ $y = {\dfrac{1}{6}x - 1}$ Substitute this expression for $y$ in the first equation. $-8x-8({\dfrac{1}{6}x - 1}) = 1$ $-8x - \dfrac{4}{3}x + 8 = 1$ Simplify by combining terms, then solve for $x$ $-\dfrac{28}{3}x + 8 = 1$ $-\dfrac{28}{3}x = -7$ $x = \dfrac{3}{4}$ Substitute $\dfrac{3}{4}$ for $x$ back into the top equation. $-8( \dfrac{3}{4})-8y = 1$ $-6-8y = 1$ $-8y = 7$ $y = -\dfrac{7}{8}$ The solution is $\enspace x = \dfrac{3}{4}, \enspace y = -\dfrac{7}{8}$.